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Proof step 9 from proof byte-law-add-pos (previous step)

j + i % 256 = i0 + 256, 0 < i % 256, 0 ≤ j, 0 ≤ i, 127 < i % 256 ⊦ i2b(i + j) = i2b(i0)

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0) subst: [i, j, i0, j0] → [i, 256, i1, j0] rotate: T spec: int-div inst: name: div-elim-pospos

Additional information

Used simplifier rules: byte-mult-256: ⊦ i2b(i * 256 + j) = i2b(j);%byte-add-256: ⊦ i2b(i + 256) = i2b(i);% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)%⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0); (elim) Used terms: i, 256, i1, j0